Inequality for Square-Summable Complex Series
Table of Contents
Some inequalities look formidable until the right decomposition makes them transparent. The conjecture below, posed by Zoltan Retkes on the Open Problem Garden in 2012 with a £10 prize attached, is one such case: once the dyadic structure of the positive integers is made explicit, the proof reduces to two classical facts.
For all $\alpha = (\alpha_1, \alpha_2, \ldots) \in \ell^2(\mathbb{C})$, $$\sum_{n \geq 1} |\alpha_n|^2 \geq \frac{6}{\pi^2} \sum_{k \geq 0} \left|, \sum_{l \geq 0} \frac{\alpha_{2^k(2l+1)}}{l+1} ,\right|^2.$$
The conjecture was confirmed by an anonymous comment on the problem page in November 2013. A self-contained proof and an extension to $\ell^p$ were subsequently published by Ibragimov and Salimova in Elemente der Mathematik 70 (2015), 79–81.
The Dyadic Decomposition #
The index $2^k(2l+1)$ running over $k \geq 0$ and $l \geq 0$ is not arbitrary: it encodes a canonical partition of the positive integers. Every $n \in \mathbb{N}^+$ factors uniquely as $$n = 2^k \cdot r, \qquad k \geq 0,\quad r \text{ odd positive},$$ where $k = v_2(n)$ is the 2-adic valuation of $n$ and $r = n/2^k$ is its odd part. Writing $r = 2l+1$ gives the bijection $\mathbb{N}_0 \times \mathbb{N}_0 \to \mathbb{N}^+$, $(k, l) \mapsto 2^k(2l+1)$. In particular the sets $$A_k = {2^k(2l+1) : l \geq 0} = {2^k, 3 \cdot 2^k, 5 \cdot 2^k, \ldots}$$ form a partition of $\mathbb{N}^+$. Explicitly: $A_0 = {1, 3, 5, 7, \ldots}$ (odd numbers), $A_1 = {2, 6, 10, 14, \ldots}$ (twice an odd number), and so on. This partition is the key structural fact behind the proof.
Proof #
The argument has two ingredients: the Basel sum $\sum_{l \geq 0}(l+1)^{-2} = \pi^2/6$, and the Cauchy–Schwarz inequality in $\ell^2(\mathbb{C})$.
Define two sequences in $\ell^2(\mathbb{C})$: $$x = \left(1,, \tfrac{1}{2},, \tfrac{1}{3},, \ldots\right), \qquad y_k = \left(\alpha_{2^k},, \alpha_{3 \cdot 2^k},, \alpha_{5 \cdot 2^k},, \ldots\right) \quad (k \geq 0).$$
The inner sum in the conjecture is exactly the $\ell^2$ inner product $\langle x, y_k \rangle$: $$\sum_{l \geq 0} \frac{\alpha_{2^k(2l+1)}}{l+1} = \langle x, y_k \rangle.$$
Step 1: Apply Cauchy–Schwarz. For each $k$,
$$|\langle x, y_k \rangle|^2 \leq |x|_2^2 \cdot |y_k|_2^2.$$
Summing over $k \geq 0$,
$$\sum _{k \geq 0} |\langle x, y _k \rangle|^2 \leq |x| _2^2 \sum _{k \geq 0} |y _k| _2^2.$$
Step 2: Evaluate using the Basel problem and the partition. The Basel problem gives $$|x| _2^2 = \sum _{l \geq 0} \frac{1}{(l+1)^2} = \frac{\pi^2}{6}.$$
Since the sets $A_k$ partition $\mathbb{N}^+$, $$\sum _{k \geq 0} |y_k|_2^2 = \sum _{k \geq 0} \sum _{l \geq 0} |\alpha _{2^k(2l+1)}|^2 = \sum _{n \geq 1} |\alpha_n|^2.$$
Combining both steps, $$\sum_{k \geq 0} \left|\sum_{l \geq 0} \frac{\alpha_{2^k(2l+1)}}{l+1}\right|^2 \leq \frac{\pi^2}{6} \sum_{n \geq 1} |\alpha_n|^2,$$ which is the inequality with the $\frac{6}{\pi^2}$ factor moved to the other side.
Sharpness of the Constant #
The constant $6/\pi^2$ is the best possible. To see this, consider the truncated sequence $\alpha^{(N)}$ defined by $\alpha^{(N)}_{2l+1} = 1/(l+1)$ for $l = 0, 1, \ldots, N-1$ and $\alpha^{(N)}_n = 0$ otherwise. Then:
- The left-hand side equals $\displaystyle\sum_{l=0}^{N-1} \frac{1}{(l+1)^2} \to \frac{\pi^2}{6}$.
- The only non-zero contribution to the right-hand side comes from $k = 0$ (since all non-zero indices are odd, i.e. in $A_0$), giving $\displaystyle\frac{6}{\pi^2}\left(\sum_{l=0}^{N-1} \frac{1}{(l+1)^2}\right)^2 \to \frac{6}{\pi^2} \cdot \frac{\pi^4}{36} = \frac{\pi^2}{6}$.
The ratio of the right-hand side to the left-hand side therefore tends to $1$ as $N \to \infty$, so no larger constant than $6/\pi^2$ can hold universally. Equality is never achieved for $\alpha \in \ell^2(\mathbb{C})\setminus{0}$ with finite norm since the limiting sequence does not belong to $\ell^2(\mathbb{C})$.
Extension to $\ell^p$ #
The Cauchy–Schwarz inequality used above is a special case of Hölder’s inequality, and the proof generalises immediately.
Let $p, q \in (1,\infty)$ with $\tfrac{1}{p} + \tfrac{1}{q} = 1$. For all $\alpha = (\alpha_1, \alpha_2, \ldots) \in \ell^p(\mathbb{C})$ and $x = (x_0, x_1, \ldots) \in \ell^q(\mathbb{C})$, $$\sum_{n \geq 1} |\alpha_n|^p \geq \left(\sum_{l \geq 0} |x_l|^q\right)^{-p/q} \sum_{k \geq 0} \left|\sum_{l \geq 0} x_l, \alpha_{2^k(2l+1)}\right|^p.$$
Retkes’s original inequality is the case $p = q = 2$ and $x_l = 1/(l+1)$, where $(\sum_{l\geq 0}|x_l|^2)^{-1} = 6/\pi^2$ by the Basel problem.
Remarks on Structure #
The role of the dyadic partition. The sets $A_k$ are the dyadic layers of $\mathbb{N}^+$: each integer sits in exactly one layer determined by its 2-adic valuation. This structure also appears in the theory of Hardy spaces, where the dyadic martingale decomposition underpins the $H^1$–BMO duality, and in wavelets, where the dyadic scaling of the real line organises the multiresolution analysis. The inequality can be read as a norm comparison between the $\ell^2$ norm and a weighted sum over dyadic layers.
Relation to the Basel problem. The constant $6/\pi^2$, the reciprocal of $\zeta(2)$, appears here because the weight sequence $1/(l+1)$ used in the inner sum is precisely the harmonic sequence, whose $\ell^2$ norm squared is $\zeta(2)$. Any other weight sequence $x \in \ell^2(\mathbb{C})$ would produce the analogous inequality with $|x|_2^{-2}$ in place of $6/\pi^2$.
The inequality as a rearrangement estimate. The right-hand side reorganises the entries of $\alpha$ by their dyadic layer and applies a weighted average within each layer. The inequality says the total $\ell^2$ energy cannot be less than $6/\pi^2$ times the energy of this rearranged, averaged version of the sequence, a quantitative statement about how averaging destroys energy.
Further Questions #
While the original conjecture is settled, several natural variants remain.
What is the sharp constant in the inequality if the dyadic partition is replaced by the partition induced by a prime $p \neq 2$, i.e. by the sets $A_k^{(p)} = {p^k m : \gcd(m, p) = 1}$? The same argument applies with $x_l = w_l$ for any weight sequence $w \in \ell^2(\mathbb{C})$, but the resulting constant depends on $|w|_2$ and the choice of weight, not on $\pi$.
The inner sum $\sum_{l \geq 0} \alpha_{2^k(2l+1)}/(l+1)$ averages the entries in layer $A_k$ with the harmonic weights. What happens if the harmonic weight $1/(l+1)$ is replaced by a weight $w(l)$ depending on the position $l$ within the layer in a more general way, for instance $w(l) = l^{-s}$ for $s > 1/2$? The sharp constant would then involve $\zeta(2s)$ instead of $\zeta(2) = \pi^2/6$.
For $p = 1$ the Ibragimov–Salimova theorem requires $q = \infty$, and the Hölder inequality takes a different form. Does an analogue of Retkes’s inequality hold for $\alpha \in \ell^1(\mathbb{C})$, and if so, what is the sharp constant?
References #
- Ibragimov, Z. O. & Salimova, D. F. (2015). On an inequality in $\ell_p(\mathbb{C})$ involving Basel problem. Elemente der Mathematik, 70(2), 79–81. https://ems.press/content/serial-article-files/45532
- Retkes, Z. (2012). Inequality for square summable complex series. Open Problem Garden. http://www.openproblemgarden.org/op/inequality_for_square_summable_complex_series
- Benko, D. & Molokach, J. (2013). The Basel problem as a rearrangement of series. College Mathematics Journal, 44(3), 171–176.
- Ritelli, D. (2013). Another proof of $\zeta(2) = \pi^2/6$ using double integrals. American Mathematical Monthly, 120(7), 642–645.